This section is from the book "The Tinman's Manual And Builder's And Mechanic's Handbook", by Isaac Ridler Butt. Also available from Amazon: The Tinman's Manual And Builder's And Mechanic's Handbook.

= When we wish to state that one quantity or number, is equal to another quantity or number, the sign of equality = is employed. Thus 3 added to 2 = 5, or 3 added to 2 is equal to 5.

+ When the sum of two quantities or numbers is to be taken, the sign plus + is placed between them. Thus 3 + 2 = 5; that is, the sum of 3 and 2 is 5. This is the sign of Addition.

- When the difference of two numbers or quantities is to be taken, the sign minus - is used, and shows that the latter number or quantity is to be taken from the former. Thus 5 - 2 = 3. This is the sign of Subtraction,

X When the product of any two numbers or quantities is to be taken, the sign into x is placed between them. Thus 3x2 = 6. This is the sign of Multiplication.

When we are to take the quotient of two quantities, the sign by ÷ is placed between them, and shows that the former is to be divided by the latter. Thus 6÷2 = 3. This is the sign of Division. But in some cases in this work, the mode of division has been, to place the dividend above a horizontal line, and the divisor below it, in the form of a vulgar fraction, thus:

Dividend / Divisor =Quotient. 6/2 =3

When the square of any number 'or quantity is to be taken, this is denoted by placing a small figure 2 above it to the right. Thus 62 shows that the square of6 is to be taken, and therefore 62 = 6 X 6 = 36.

When we wish to show that the square root of any number or quantity is to be taken, this is denoted by placing the radical sign √ before it. Thus √36 shows that the square root of 36 ought to be taken, hence √36 = 6.

The common marks of proportion are also used, viz.,:::: as 3:6 :: 4: 8, being read 3 is to 6 as 4 is to 8.

The application of these signs to the expression of rules is exceedingly simple. Thus, connected with the circle we have the following rules:

1st. The circumference of a circle will be found by multiplying the diameter by 3T416.

2d. The diameter of a circle may be found by dividing the circumference by 3T416.

3d. The area of a circle may be found by multiplying the half of the diameter, by the half of the circumference, or by multiplying together the diameter and circumference, and dividing the product by 4, or by squaring the diameter and multiplying by 7864.

Now all these rules may be thus expressed:

1st. diameter X 3.1416 = circumference.

circumference/ 3.1416 = diameter ,.

2d diameter/2 X circumference/2 =area 3d. diameter X circumference/ 4 =area or, diameter2 X . 7851 = area.

Practical Geometry is an important branch of knowledge to all who are in any way engaged in the art of building. The workman, as well as the designer, requires its aid; and unless he is acquainted with some of the leading principles of the science, he will frequently feel an uncertainty as to the results he may deduce from the problems which are presented to his notice.

Problem I.

To inscribe an Equilateral Triangle within a given Circle. Let a b c be a circle; it is required to draw within it a triangle whose sides are equal to one another. Commencing from any point A, mark on the circumference of the circle a series of spaces equal to the radius of the circle, of which there will be six, and draw the arcs a d d b, etc. Then join every alternate point as a b, b c, c a, and the several lines will together form an equilateral triangle.

Fig. 1.

Problem II.

Within a given Circle to inscribe a Square. Let A b c d be the given circle, it is required to draw a square within it. Draw the diameters a b, c d, at right angles to each other; or, in other words, draw the diameter a b, and form a perpendicular bisecting it. Then join the points a c, c b, b d, d a, and the figure a b c d is a square formed within a given circle.

Fig. 2.

Problem III.

Within a given Circle to inscribe a regular Pentagon; that is, a Polygon of five Sides.

Let a b C D be a circle in which it is required to draw a pentagon.

Fig. 3.

Draw a diameter a D, and perpendicular to it another diameter. Then divide O b into two equal parts in the point e, and join c e; and with e as a centre, and the radius c E, draw the arc c f, cutting a o in f: and, with c as a centre, and the same radius, describe the arc f g; the arcs c f, g f intersect each other in the point f, and the arc G f intersects the circumference of the circle in the point G. Join the points c and G, and that line will be a side of the pentagon to be drawn. Mark off within the circumference the same space, and join the points a H, HI, IK K, K c, and the figure that is formed is a pentagon.

Problem IV.

Within a given Circle to describe a regular Hexagon; that is to say, a Polygon of six equal Sides.

Let a b c be the given circle, and O the centre. With the radius of the circle divide it into parts, of which there will be six, and connect the points a d, d b, etc, and the figure adbecf will be a regular hexagon.

Fig. 4.

Problem V.

To cut off the Corners of a given Square, so as to form a regular

Octagon.

Let a b c d be the given square. Draw the two diagonal lines

Fig. 5.

A c, and b d, crossing each other in o. Then, with the radius A O, that is, half the diagonal, and with a as a centre, describe the arc e f, cutting the sides of the square in e and f; then, from b as a centre, describe the arc g h; and in like manner from c and d describe the arcs I k and l m. Draw the lines l g, f i, h m, and k e, and these, with the parts of the given square G f, i h, m k, and e l, form the octagon required.

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